SOLUTION: How do I solve this problem? the first two terms of an arithmetic sequence are a(base1)=2 and a(base2)=4. what is the a(base10)?

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Question 175215: How do I solve this problem? the first two terms of an arithmetic sequence are a(base1)=2 and a(base2)=4. what is the a(base10)?
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
the formula for the nth term of a geometric sequence is:
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a%5Bn%5D+=+a%5B1%5D%2Ar%5E%28n-1%29
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you are given a%5B1%5D and a%5B2%5D.
you use them to find r as follows:
a%5B2%5D+=+a%5B1%5D%2Ar%5E%282-1%29 which becomes a%5B2%5D+=+a%5B1%5D%2Ar%5E1
substituting 4 for a%5B2%5D and 2 for a%5B1%5D you get:
4 = 2*r^1 = 2*r
solving for r, you get
r = 2
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you now have r, and you can substitute in the general equation as follows:
a%5Bn%5D+=+a%5B1%5D%2Ar%5E%28n-1%29
substitute 2 for r, and 2 for a%5B1%5D and you get:
a%5Bn%5D+=+2%2A2%5E%2810-1%29 = 2*2^9 = 2*512 = 1024.
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the 10th term in the sequence is 1024.
you can test this by doing it the hard way (each term is done separately) as follows:
a(1) = 2
a(2) = 4
a(3) = 8
a(4) = 16
a(5) = 32
a(6) = 64
a(7) = 128
a(8) = 256
a(9) = 512
a(10) = 1024
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each succeeding number is multiplied by the ratio.
you do that 9 times which is the same as multiplying the original number by 2^9.